$f(x) = \begin{cases} -8 & \text{if } x = 1 \\ -2x^{2}-2 & \text{otherwise} \end{cases}$ What is the range of $f(x)$ ?
Answer: First consider the behavior for $x \ne 1$ Consider the range of $-2x^{2}$ The range of $x^2$ is $\{\, y \mid y \ge 0 \,\}$ Multiplying by $-2$ flips the range to $\{\, y \mid y \le 0 \,\}$ To get $-2x^{2}-2$ , we subtract $2$ So the range becomes: $\{\, y \mid y ≤ -2 \,\}$ If $x = 1$, then $f(x) = -8$. Since $-8 ≤ -2$, the range is still $\{\, y \mid y ≤ -2 \,\}$.